Explanation Given information: The diameter of the shaft is d = 100 mm . Assumptions: Young’s modulus of the 2014-T6 aluminum shaft is E = 73.1 × 10 6 kN/m 2 . Explanation: Virtual-work equation for the entire beam is: 1 ⋅ Δ = ∫ 0 L m M E I d x (1) Here, 1 is external virtual unit load acting on the beam in the direction of Δ , Δ is displacement caused by the real loads on the beam, m is internal virtual moment in the beam, expressed as a function of x and caused by the external virtual unit load, M is internal moment in the beam, expressed as a function of x and caused by the real loads, E is modulus of elasticity of a member, and I is moment of inertia of the cross-sectional area about the neutral axis. Internal moment in the beam ( M ): Determine the reactions: Entire shaft: Show the free-body diagram of the entire shaft as in Figure 1. Moment about point A : Determine the vertical reaction at point B by taking moment about point A . ∑ M A = 0 B y ( 3 ) − 8 ( 2 ) − 8 ( 1 ) = 0 (2) Along the vertical direction: Determine the vertical reaction at point A by resolving the vertical component of forces. ∑ F y = 0 A y + B y − 8 − 8 = 0 (3) Show the calculation of values as follows: Conclusion: Solve Equation (2). B y ( 3 ) − 16 − 8 = 0 B y = 8 kN Substitute 8 kN for B y in Equation (3). A y + 8 − 8 − 8 = 0 A y = 8 kN Region 0 ≤ x < 1 m : Show the free body diagram of the shaft for the region 0 ≤ x < 1 m as in Figure 2. Moment about the section: Determine the moment at the section by taking moment about the section. ∑ M x 1 = 0 M 1 − A y ( x 1 ) = 0 (4) Substitute 8 kN for A y Equation (4). M 1 − 8 ( x 1 ) = 0 M 1 = 8 x 1 Region 1 m < x < 1.5 m : Show the free body diagram of the shaft for the region 1 m < x < 1.5 m as in Figure 3. Moment about the section: Determine the moment at the section by taking moment about the section. ∑ M x 2 = 0 M 2 − A y ( 1 + x 2 ) + 8 ( x 2 ) = 0 (5) Substitute8 kN for A y in Equation (5). M 2 − 8 ( 1 + x 2 ) + 8 ( x 2 ) = 0 M 2 − 8 − 8 x 2 + 8 x 2 = 0 M 2 = 8 kN-m Internal virtual moment in the beam ( m ): Apply a unit moment at point A and calculate the moments. Determine the reactions: Entire shaft: Show the free-body diagram of the entire shaft as in Figure 4. Moment about point A : Determine the vertical reaction at point B by taking moment about point A . ∑ M A = 0 − 1 ( 1.5 ) + B y ( 3 ) = 0 (6) Along the vertical direction: Determine the vertical reaction at point A by resolving the vertical component of forces. ∑ F y = 0 A y + B y − 1 = 0 (7) Show the calculation of values as follows: Conclusion: Solve Equation (6). 3 B y = 1.5 B y = 0.5 Substitute 0.5 for B y in Equation (7). A + 0.5 − 1 = 0 A y = 0.5 Region 0 ≤ x < 1 m : Show the free body diagram of the shaft for the region 0 ≤ x < 1 m as in Figure 5. Moment about the section: Determine the moment at the section by taking moment about the section. ∑ M x 1 = 0 m 1 − A y ( x 1 ) = 0 (8) Substitute 0.5for A y in Equation (8). m 1 − 0

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ISBN: 2818440034374

Author: HIBBELER

Publisher: PEARSON

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Slope and Deflection

The slope in a deflected beam is described as an angle in radians made by the tangent of the section with the original axis of the beam. The deflection at any point on the axis of the beam is defined as the lateral displacement of the point before and af…

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Chapter 14.7, Problem 14.104P

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The slope at A of the 2014-T6 aluminum shaft.

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Chapter 14.7, Problem 14.103P

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The slope at A of the 2014-T6 aluminum shaft.

Solution Summary: The author shows the free-body diagram of the 2014-T6 aluminum shaft as in Figure 1. Determine the vertical reaction at point A by resolving vertical component of forces.

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The slope at A of the 2014-T6 aluminum shaft.

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